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Simple question

  1. Oct 7, 2004 #1
    What does it mean if an equation is linear?

    Like what is a linear differential equation?
     
  2. jcsd
  3. Oct 7, 2004 #2

    JasonRox

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    A linear equation is a straight line where m is defined. (m is the slope)

    It is in the form y=mx+b, where m is the slope and b is the y-intercept.

    This second part is a guess.

    A linear differential equation I "THINK" is a polynomial to the second degree.

    I think this because the derivative of:
    [tex]y=Nx^2+Mx+W[/tex]
    is
    [tex]\frac{dy}{dx}=2Nx+M[/tex]

    ...which is linear.

    2N is the slope of the derivative and M is the y-intercept.
     
  4. Oct 8, 2004 #3

    cepheid

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    No.

    Remember that a differential equation is an equation that contains a function y and one or more of its derivatives. The equation need only be linear in y, but not necessarily in the independent variable (x). The general form of a linear first-order differential equation is:

    [tex] \frac{dy}{dx} + p(x)y = g(x) [/tex]

    Where [itex] p(x) \ \ \text{and} \ \ g(x) [/itex] are functions of x. So if I write:

    [tex] \frac{dy}{dx} + (\cos{x})y = e^x [/tex]

    I guarantee you that the solution is not a second-degree polynomial in x! I seem to remember there being a great tutorial thread on D.E.'s in that subforum.

    Jason, I guess you can think of the equation you wrote as the simplest case of a linear first order d.e., in which p(x) = 0. So we're left with:

    [tex] \frac{dy}{dx} = g(x) [/tex]

    The solution can be found simply by integrating in this case. Not so simple with the previous example.
     
  5. Oct 8, 2004 #4

    HallsofIvy

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    The crucial point about a "linear differential equation", indeed about "linear" problems in general, is that we can combine two solutions to make a third solution.

    If y1 and y2 both satisfy the equation a(x)y"+ b(x)y'+ c(x)y= 0 then any "linear combination" of them, py1+ qy2, where p and q are numbers, does also:
    a(x)(py1+ qy2)"+ b(x)(py1+ qy2)'+ c(x)(py1+ qy2)=
    a(x)py1"+ a(x)qy2"+b(x)py1'+ b(x)qy2'+ c(x)py1+ c(x)qy2[/sub=

    p(a(x)y1"+ b(x)y1'+ c(x)y1)+ q(a(x)y2"+ b(x)y2'+ c(x)y2=

    p(0)+ q(0)= 0.

    Similarly, if f(x)= ax, then f(nx+my)= a(nx+my)= n(ax)+ m(ay)= nf(x)+ mf(y).

    If, however, f(x)= x2, then f(x+y)= (x+ y)2= x2+ 2xy+ y2. The fact that f is NOT linear means that that term 2xy in which the two solutions x and y "interfere" with one another.
     
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